y=x(x^(2)+b^(2))^(- 1/2 ) get the derivative

To find the derivative of the given function \( y = x(x^2 + b^2)^{-\frac{1}{2}} \), we need to apply both the product rule and the chain rule.

### Step 1: Identify the components of the function
Let:
\[
y = u \cdot v
\]
where
\[
u = x \quad \text{and} \quad v = (x^2 + b^2)^{-\frac{1}{2}}
\]

### Step 2: Find the derivatives of \( u \) and \( v \)
First, differentiate \( u = x \) with respect to \( x \):
\[
u' = \frac{du}{dx} = 1
\]

Next, differentiate \( v = (x^2 + b^2)^{-\frac{1}{2}} \) using the chain rule:
\[
v' = \frac{d}{dx} \left( (x^2 + b^2)^{-\frac{1}{2}} \right)
\]
Let \( w = x^2 + b^2 \), so \( v = w^{-\frac{1}{2}} \). Then,
\[
v' = -\frac{1}{2} w^{-\frac{3}{2}} \cdot \frac{dw}{dx}
\]
Since \( \frac{dw}{dx} = 2x \), we have:
\[
v' = -\frac{1}{2} \cdot (x^2 + b^2)^{-\frac{3}{2}} \cdot 2x = -\frac{x}{(x^2 + b^2)^{\frac{3}{2}}}
\]

### Step 3: Apply the product rule
Now apply the product rule to find \( y' \):
\[
y' = u'v + uv'
\]
Substitute the values for \( u' \), \( v \), \( u \), and \( v' \):
\[
y' = (1) \cdot (x^2 + b^2)^{-\frac{1}{2}} + x \cdot \left(-\frac{x}{(x^2 + b^2)^{\frac{3}{2}}}\right)
\]
Simplify:
\[
y' = (x^2 + b^2)^{-\frac{1}{2}} - \frac{x^2}{(x^2 + b^2)^{\frac{3}{2}}}
\]

### Step 4: Combine the terms
Factor out the common factor \( (x^2 + b^2)^{-\frac{3}{2}} \):
\[
y' = \frac{(x^2 + b^2) - x^2}{(x^2 + b^2)^{\frac{3}{2}}}
\]
Simplify the numerator:
\[
y' = \frac{b^2}{(x^2 + b^2)^{\frac{3}{2}}}
\]

### Final Answer
The derivative of the function \( y = x(x^2 + b^2)^{-\frac{1}{2}} \) is:

\[
y' = \frac{b^2}{(x^2 + b^2)^{\frac{3}{2}}}
\]

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Would you like further details or have any questions?

Here are five related questions:
1. What happens to the derivative as \( x \) approaches 0?
2. How does the derivative behave for large values of \( x \)?
3. How would the derivative change if the exponent in the original function was different?
4. Can you determine the second derivative of the function?
5. How would the derivative be different if the function were \( y = (x^2 + b^2)^{-\frac{1}{2}} \) without the \( x \) factor?

**Tip:** When dealing with products of functions, always consider whether the product rule is necessary, and use the chain rule for any composite functions.

Learn how to find the derivative of y = x(x^2 + b^2)^{-1/2} using the product rule and chain rule. This problem covers derivatives of composite functions and is suitable for college-level calculus students.

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